Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 209

( 2 , 1) for ( a, b, c ) (later, we'll see a faster way to eliminate them). Only

the combinations

− 2 , −

( a, b, c )=(1 , 1 , 1) , ( − 1 , − 2 , 2) , (2 , 2 , 1) , ( − 2 , − 1 , 2)

remain. As we'll see below, these four combinations correspond to the four
points that we already know about, namely,

∞ , (0 , 0) , (2 , 0) , ( − 2 , 0)

(this requires some explanation, which will be given later). As we'll see later,
the fact that we eliminated all combinations except those coming from known
points implies that we have found all points, except possibly points of odd
order, on the curve. The Lutz-Nagell theorem, or reduction mod 5 and 7
(see Theorem 8.9), shows that there are no nontrivial points of odd order.
Therefore, we have found all rational points on E :

E ( Q )= {∞, (0 , 0) , (2 , 0) , ( − 2 , 0) }.

The calculations of the example generalize to elliptic curves E of the form

y 2 =( x − e 1 )( x − e 2 )( x − e 3 )

with e 1 ,e 2 ,e 3 ∈ Z and e i = j . In fact, they extend to even more

general situations. If e i ∈ Q but e i ∈ Z , then a change of variables transforms
the equation to one with e i ∈ Z , so this situation gives nothing new. However,
if e i ∈ Q , the method still applies. In order to keep the discussion elementary,
we'll not consider this case, though we'll say a few things about it later.
Assuming that x, y ∈ Q ,write

= e j when i

x − e 1 = au 2
x − e 2 = bv 2
x − e 3 = cw 2

with rational numbers a, b, c, u, v, w .Then y 2 = abc ( uvw ) 2 ,so

abc is a square.

By adjusting u, v, w , we may assume that a, b, c are squarefree integers.

PROPOSITION 8.13
Let

S = {p | p isprimeand p| ( e 1 − e 2 )( e 1 − e 3 )( e 2 − e 3 ) }.

If p isaprimeand p|abc ,then p ∈ S .